Question on Indefinite Integration: $\int\frac{2x^{12}+5x^9}{\left(x^5+x^3+1\right)^3}\,\mathrm{d}x$

Question

Give me some hints to start with this problem: $${\displaystyle\int}\dfrac{2x^{12}+5x^9}{\left(x^5+x^3+1\right)^3}\,\mathrm{d}x$$

Answer 1

Let $x=1/t$. Then $dx=-dt/t^2$. Our integral becomes $$ \int -\frac{2t+5t^4}{(1+t^2+t^5)^3}\,dt.$$

Answer 2

HINT: $$x^5+x^3+1=x^5(1+x^{-2}+x^{-5})$$

Choose $1+x^{-2}+x^{-5}=u$

Answer 3

Hint

Take $x^5$ common from denominator so that it's cube it will come out as $15$ and take $x^{15}$ common from the numerator. Put the remaining denominator as $t$ numerator becomes $dt$. Hope you can do the rest.

Answer 4

Because of the cube in denominator, you can say that $$\int \frac{2x^{12}+5x^9}{\left(x^5+x^3+1\right)^3}\,dx=\frac{P_n(x)}{(x^5+x^3+1)^2}$$ where $P_n(x)$ is a polynomial fo degree $n$.

Now differentiate both sides to get $$\frac{2x^{12}+5x^9}{\left(x^5+x^3+1\right)^3}=\frac{\left(x^5+x^3+1\right) P_n'(x)-2 x^2 \left(5 x^2+3\right) P_n(x)}{\left(x^5+x^3+1\right)^3}$$ So, the right hand side is of degree $n+4$; then, since the left hand side is of degree $12$, this implies $n=8$.

So, set $P_8(x)=\sum_{i=0}^8 a_i x^i$; replacing and comparing the terms of same power then gives the $a_i$'s.

Edit

This works but it is just junk when compared to the so beautiful solution given by André Nicolas.